Lottery Probability

Well firstly I would just like to say, that I don't only want to work out the probability, I also want to make some exceptions. We will use a typical lottery, 6 numbers are drawn from a range of 49 and if the 6 numbers on a ticket match the numbers drawn, the ticket holder is a jackpot winner.

I understand that the numbers 1,2,3,4,5,6 have the same probablility as any other six numbers. BUT! I want to exclude all consecutive numbers so I would like to calculate the probability of winning the lottery, but I don't want to take into account any consecutive numbers. For example:

1, 13, 23, 45, 46, 7

I know that this sounds stupid, but I just want to exclude them.

This is how far I've got... (Skip red if you know)

Starting with a bag of 49 differently-numbered lottery balls, there is clearly a 1 in 49 chance of predicting the number of the first ball selected from the bag. When the second number is picked there are now only 48 balls so there is a 1 in 48 chance and so on.

49 x 48 x 47 x 46 x 45 x 44 = 10,068,347,520

Next the order in which the numbers are selected doesn't matter, so...

1 x 2 x 3 x 4 x 5 x 6 = 720

10,068,347,520 / 720 = 13,983,816

So there is a 1 in 13,983,816 chance of winning?

Ok now this is the bit I am struggling with! How do I now ignore any set that has 2 or more consecutive numbers? Any help would be great, it's just something I am interested. I guess it's the fact that my mind is telling me that consecutive numbers isn't going to happen frequently, and It'll make my chances look mor elikely if I ignore them, and will make me think I'm going to win. 5% of my winnings to the first person to help

Letís discuss the commonly understood lottery game.
If we draw out six numbered balls from forty-nine, order makes no difference.
Only content matters. The combination {2,4,27,33,41,42} is the same as {41,2,33,4,27,42}. If the first is the winning combination then so is the second. is the number of possible winning combinations.

However, if you note that example contains a pair of consecutive numbers, 41 & 42. That is what you do not want. Well sometime ago some players, in I think Louisiana, noticed that quite often winning combinations did have consecutive pairs. They want to file a legal complaint. But as it turns out it not that unreasonable for it to happen.
We can think of that a string of six ones and 43 zeros as one ticket. Any arrangement of that string could represent a winning combination. So your question comes down to how many rearrangements of that bit string have no two consecutive ones. The answer is the zeros create 44 places to put the ones. So for this particular set of numbers there almost a 50/50 chance a winning ticket will have consecutive entries.

Now of course, if your own lottery game turns on the order in which the balls come out of the hopper, then a totally different analysis must be done. But in the usual lottery order does not matter.

The order does not matter, what I want to do it remove all cases of 2 or more consecutive numbers in a set and remove that and then get a probibility.

I suppose that you did not read my response very carefully.
If there is a pool of N numbers from which k are selected to determine a winning ticket, then the probability that two consecutive integers are not chosen is
I have generated a table to show the probabilities for several different values of N & k.
I have highlighted your particular case of N=49 and k=6.